The Vee - 8 months ago 33

C++ Question

This is my first encounter with the visitor design pattern for double dispatch. In my scenario objects derived from a common base class can interact somehow with each other, so the visitor and the visitee are from the same set. However, for practical reasons, my goal is to completely separate the *logic* from the *application*. In particular, I want to avoid **at all costs** the base class to be aware of what derived classes can be using it. There will be different independent application layers approaching the same base within one application.

The problem is that a

`virtual`

`/*** Generic ***/`

template<class...>

struct Visit { };

template<class... Derived>

struct Base : Visit<Derived...> {

virtual int accept(Base*) = 0;

};

// specializations for different numbers...

template<class X>

struct Visit<X> {

virtual int visit(X*) = 0;

};

template<class X, class Y>

struct Visit<X, Y> {

virtual int visit(X*) = 0;

virtual int visit(Y*) = 0;

};

// and more for 3, 4, ...

/*** Application ***/

struct D1;

struct D2;

using A = Base<D1, D2>; // <--- the goal is to keep this simple

struct D1 : A {

int accept(A* a) { return a->visit(this); }

int visit(D1*) { return 1; }

int visit(D2*) { return 2; }

};

struct D2 : A {

int accept(A* a) { return a->visit(this); }

int visit(D1*) { return 3; }

int visit(D2*) { return 4; }

};

int main() {

A* d1 = new D1();

A* d2 = new D2();

return d2->accept(d1); // expected: 2

}

This works and satisfies most of the criteria except the last one. The maximum number of possible derived classes needs to be known in advance and hard-coded in the

`Visit`

I was wondering whether something cleaner along the lines of

`template<class X>`

struct InjectVisit {

virtual int visit(X*) = 0;

};

template<class... Derived>

struct Base : InjectVisit<Derived>... {

virtual int accept(Base*) = 0;

};

(replacing the

`Visit`

Overload resolution only selects a base template (or a nontemplate function, if one is available). Only after it's been decided which base template is going to be selected, and that choice is locked in, will the compiler look around to see if there happens to be a suitable specialization of that template available, and if so that specialization will get used.

Since each of the injected

`visit(X*)`

`InjectVisit`

I tried to adapt the second half of this answer but it won't work if

`D1`

`D2`

`dynamic_cast`

Currently I am stuck on a middle way in which the base class of

`Base`

`Visit`

Answer

Since each of the injected

`visit(X*)`

comes from a different template instantiation of`InjectVisit`

, they won't compete against each other, leading to ambiguity errors (even though only exactly one of them could be used at any point).

You could use following `using`

trick:

```
#include <iostream>
void println(const char *s)
{
using namespace std;
cout << s << endl;
}
template<typename X>
struct InjectVisit
{
virtual void visit(X*) = 0;
};
template<typename Head, typename ...Tail>
struct VirtualChain : InjectVisit<Head>, VirtualChain<Tail...>
{
using InjectVisit<Head>::visit;
using VirtualChain<Tail...>::visit;
};
template<typename Head>
struct VirtualChain<Head> : InjectVisit<Head>
{
using InjectVisit<Head>::visit;
};
template<typename ...List>
struct Base : VirtualChain<List...>
{
virtual void accept(Base*) = 0;
};
/****************************************************************/
struct D1;
struct D2;
using ConcreteBase = Base<D1, D2>;
struct D1 : ConcreteBase
{
virtual void accept(ConcreteBase* visitor) { visitor->visit(this); }
virtual void visit(D1*) { println("D1 visited by D1"); }
virtual void visit(D2*) { println("D2 visited by D1"); }
};
struct D2 : ConcreteBase
{
virtual void accept(ConcreteBase* visitor) { visitor->visit(this); }
virtual void visit(D1*) { println("D1 visited by D2"); }
virtual void visit(D2*) { println("D2 visited by D2"); }
};
int main()
{
ConcreteBase* d1 = new D1();
ConcreteBase* d2 = new D2();
d1->accept(d2);
d2->accept(d2);
}
```

Output is:

```
D1 visited by D2
D2 visited by D2
```

Source (Stackoverflow)